For `x in R, x ne -1`, if `(1 + x)^(2016) = sum_(i = 0)^(2016) a_(i)x^(i)`, then `a_(17)` is equal to :

Options

A. `(2017!)/(17!2000!)`
B. `(2016!)/(17! 1999!)`
C. `(2016!)/(16!)`
D. `(2017!)/(2000!)`

Answer

To find the coefficient \( a_{17} \) in the expansion of \( (1 + x)^{2016} \), we can use the Binomial Theorem. The Binomial Theorem states that:

\[ (1 + x)^n = \sum_{i=0}^{n} \binom{n}{i} x^i \]

where \( \binom{n}{i} \) is the binomial coefficient, which can be calculated as:

\[ \binom{n}{i} = \frac{n!}{i!(n-i)!} \]

In this case, we have \( n = 2016 \) and we want to find the coefficient of \( x^{17} \), which corresponds to \( i = 17 \).

### Step 1: Identify \( n \) and \( i \)

Here, \( n = 2016 \) and \( i = 17 \).

### Step 2: Calculate the binomial coefficient \( \binom{2016}{17} \)

Using the formula for the binomial coefficient:

\[ \binom{2016}{17} = \frac{2016!}{17!(2016 - 17)!} = \frac{2016!}{17! \cdot 1999!} \]

### Step 3: Simplify the expression

We can simplify \( \frac{2016!}{1999!} \):

\[ \frac{2016!}{1999!} = 2016 \times 2015 \times 2014 \times \ldots \times 2000 \]

Thus, we can write:

\[ \binom{2016}{17} = \frac{2016 \times 2015 \times 2014 \times \ldots \times 2000}{17!} \]

### Step 4: Calculate \( a_{17} \)

Now, we have:

\[ a_{17} = \binom{2016}{17} \]

### Final Result

The value of \( a_{17} \) is:

\[ a_{17} = \frac{2016 \times 2015 \times 2014 \times \ldots \times 2000}{17!} \]